Maximum area of triangle

How would you find the largest possible area of a triangle in the first quadrant bounded by the two axis and any tangent of the curve $e^{-5 x}$?

I suppose the solution involves formulating a function for the area and using the derivative to find the maximum value. But I struggle to determine the function..

Solution 1:

For a given function $f(x)$ and a point $(x_0, y_0): y_0 = f(x_0)$, the tangent line is defined as

$$ y = f(x_0) + f'(x_0)(x-x_0). $$

Given that $f(x) = e^{-5x}$, the derivative $f'(x) = -5e^{-5x}$, and we have a tangent line

$$ y = e^{-5x_0}-5e^{-5x_0}(x-x_0). $$

This line intersects

• $y$-axis at $\left(x=0, y = e^{-5x_0}+5e^{-5x_0}x_0\right)$;
• $x$-axis at $\left(x=\frac{1}{5}+x_0, y = 0\right)$.

Then, the triangle of interest is a right triangle with two legs:

• $a = e^{-5x_0}+5e^{-5x_0}x_0 = e^{-5x_0}\left(1+5x_0\right)$,
• $b = \frac{1}{5}+x_0$.

$x_0 > 0$ since the triangle belongs to the first quadrant.

The area of the triangle equals to

$$ S(x_0) = \frac{e^{-5x_0}}{2}\left(1+5x_0\right)\left(\frac{1}{5}+x_0\right) = \frac{e^{-5x_0}}{2}\left(\frac{1}{5}+2x_0+5x_0^2\right) $$

And the problem now is to find a maximum of $S(x_0)$, $x_0 > 0$.

$$ \begin{aligned} S'(x_0)&=\frac{-5e^{-5x_0}}{2}\left(\frac{1}{5}+2x_0+5x_0^2\right)+ \frac{e^{-5x_0}}{2}\left(2+10x_0\right) = \\ &= \frac{e^{-5x_0}}{2}\left(1-25x_0^2\right) \\\\ S'(x_0) &= 0 \Leftrightarrow 1-25x_0^2 = 0 \Leftrightarrow x_0 = \pm \frac{1}{5}, \end{aligned} $$

and $x_0 = \frac{1}{5}$ is a point where $S(x_0)$ is maximal.

So, the maximum area equals to

$$ \begin{aligned} S\left(\frac{1}{5}\right) = \frac{e^{-1}}{2}\left(\frac{1}{5}+\frac{2}{5}+5\left(\frac{1}{5}\right)^2\right) = \frac{e^{-1}}{2}\frac{4}{5} = \frac{2}{5}e^{-1}. \end{aligned} $$